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Definition: If $X$ and $Y$ are topological spaces. The product topology on $X \times Y$ is the topology having basis the collection $\mathcal{B}$ of all sets of the form $U \times V$, where $U$ is an open set of $X$ and $V$ is an open set of $Y$

I've taken this definition from Munkres: Topology - A First Course and the notation he uses at times can be highly confusing. For example he uses $x \times y$ to denote an ordered pair $(x, y)$.

Now I'm not sure in the definition whether $U \times V$ is denoting the cartesian product of the two sets, or the ordered pair $(U, V)$

So which of the following is the correct basis for the product topology as stated in the definition above, given that $\mathcal{T_X}$ is the topology on $X$ and $\mathcal{T_Y}$ is the topology on $Y$?

$$\mathcal{B} = \left\{U \times V \ \middle| \ U \in \mathcal{T_X} \ \text{ and } \ V \in \mathcal{T_Y}\right\}$$ $$\mathcal{B'} = \left\{(U, V) \ \middle| \ U \in \mathcal{T_X} \ \text{ and } \ V \in \mathcal{T_Y}\right\} = \mathcal{T_X} \times \mathcal{T_Y}$$

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    $\begingroup$ $U\times V$ here is the Cartesian product of $U$ and $V$. (And yes, his notation for ordered pairs is abominable.) Your first $\mathscr{B}$ is correct. $\endgroup$ – Brian M. Scott Jan 6 '17 at 14:25
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As already said in comments, the basis consists of Cartesian products $U\times V$, where $U$ is open in $X$ and $V$ is open in $Y$.

You can notice that that other possibility does not make sense, since sets from $\mathcal B$ must be subsets of $X\times Y$.


Product of finitely many topological spaces is defined in similar way as for two spaces. But perhaps it is worth to add a word of warning that if you define product of arbitrarily many spaces, then you have to be a bit more careful. (If you simply take products of open sets, you would get box topology, which has different properties from product topology.)

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